1. Problem: In how many ways can 10 classroom keys be arranged in a circular key chain? Solution: Arranging $n$ distinct objects in a circle has $(n-1)!$ ways. So, number of ways = $ (10-1)! = 9! = 362,880 $. 2. Problem: In how many ways can 10 candles of different colors be arranged in a row? Solution: Arranging $n$ distinct objects in a row has $n!$ ways. Number of ways = $10! = 3,628,800$. 3. Problem: In how many ways can 10 candles of different colors be arranged in a circular candle holder? Solution: Number of circular arrangements of $n$ objects = $(n - 1)!$. Number of ways = $9! = 362,880$. 4. Problem: In how many ways can 12 members select a president, vice-president, secretary, and treasurer? Solution: Number of ways = $P(12,4) = \frac{12!}{(12-4)!} = \frac{12!}{8!} = 12\times11\times10\times9 = 11,880$. 5. Problem: Two parents and 3 children are seated in a row; parents must not sit together. Find number of arrangements. Solution: Total permutations of 5 people = $5! = 120$. Treat parents as one unit: number of arrangements with parents together = $4! \times 2! = 24 \times 2 = 48$. So, parents not together = $120 - 48 = 72$ ways. 6. Problem: Teacher Matilda wants to put 4 Grammar, 5 Science, and 6 Math books on a shelf so that books of the same subject are together. Solution: Treat each subject group as one block: 3 blocks can be arranged in $3! = 6$ ways. Within the blocks, books can be arranged as: Grammar: $4! = 24$ ways Science: $5! = 120$ ways Math: $6! = 720$ ways Total ways = $3! \times 4! \times 5! \times 6! = 6 \times 24 \times 120 \times 720 = 1,244,160$ ways. 7. Problem: How many four-letter arrangements are possible from the letters T, U, E, S, D, A, Y? Solution: Number of letters = 7, choosing and arranging 4 distinct letters: Number of arrangements = $P(7,4) = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$. 8. Problem: How many distinct nine-digit numbers can be formed using digits 1, 2, 1, 2, 3, 1, 2, 3, 4? Solution: Count digits: 1 appears 3 times, 2 appears 3 times, 3 appears 2 times, 4 appears 1 time. Total permutations = $\frac{9!}{3!\times 3! \times 2! \times 1!} = \frac{362,880}{6 \times 6 \times 2} = \frac{362,880}{72} = 5,040$. 9. Problem: In how many ways can 5 men and 5 ladies sit around a table if no 2 men and no 2 ladies sit together? Solution: They must alternate men and ladies. Fix men's seats: number of ways to arrange men in circle = $(5-1)! = 4! = 24$. Arrange ladies in remaining seats: $5! = 120$. Total = $24 \times 120 = 2,880$ ways. 10. Problem: Top 3 of 12 champions get gold, silver, bronze medals. How many winning orders? Solution: Number of permutations = $P(12,3) = 12 \times 11 \times 10 = 1,320$. 11. Problem: Arrange marshals - 8 of 12 Team A on left, 9 of 13 Team B center, 6 of 10 Team C right side. Solution: Each group's arrangements: Team A: $8! = 40,320$ Team B: $9! = 362,880$ Team C: $6! = 720$ Total ways = $8! \times 9! \times 6! = 40,320 \times 362,880 \times 720 = 10,543,867,546,880$. 12. Problem: Calculate $11!$ (BALLESTEROS P11) Solution: $11! = 39,916,800$. 13. Problem: Calculate $12!$ (SANGKATAUHAN P12) Solution: $12! = 479,001,600$. 14. Problem: Calculate $11!$ (MANGANTERAN P11) Solution: $11! = 39,916,800$. Final answers summary: 1) $9! = 362,880$ 2) $10! = 3,628,800$ 3) $9! = 362,880$ 4) $11,880$ 5) $72$ 6) $1,244,160$ 7) $840$ 8) $5,040$ 9) $2,880$ 10) $1,320$ 11) $10,543,867,546,880$ 12) $39,916,800$ 13) $479,001,600$ 14) $39,916,800$